Ask your own question, for FREE!
Mathematics 54 Online
OpenStudy (anonymous):

evaluate the sum of the finite geometric series. 1+2+4+8+...+128

jimthompson5910 (jim_thompson5910):

S = a*((1 - r^(n+1))/(1-r)) S = 1*((1 - 2^(7+1))/(1-2)) S = 1*((1 - 2^8)/(1-2)) S = 1*((1 - 256)/(1-2)) S = 1*((-255)/(-1)) S = 255 So 1+2+4+8+...+128 = 255

OpenStudy (anonymous):

what is the original equation? So i can use it on the other problems

jimthompson5910 (jim_thompson5910):

If you have the geometric sequence, a[n] = a*r^(n-1), then the sum from the first term 'a' to the nth term a*r^(n-1) is S = a*((1 - r^n)/(1-r))

OpenStudy (saifoo.khan):

Ops. i read infinite. lol

OpenStudy (anonymous):

for powers of two you can double the last number and subtract one \[1+2+2^2+...+2^n=2^{n+1}-1\] so in your case you can take 128, double it and get 256, subtract 1 get 255

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!